Problem: Let $g(x)=(2x^3+x^2-6x)^{^{{\scriptsize\dfrac23}}}$. Find $g'(2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $-\dfrac{88}{3}$ (Choice C) C $\dfrac13$ (Choice D) D $\dfrac{22}{3}$
Let's start by finding the expression for $g'(x)$. Then, we can evaluate it at $x=2$. $g$ is a power function with a rational exponent, but its argument isn't simply $x$. Therefore, it's a composite power function. In other words, suppose $u(x)=2x^3+x^2-6x$, then $g(x)=[{u(x)}]^{^{{\scriptsize\dfrac23}}}$. $g'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[[{u(x)}]^{^{{\scriptsize\dfrac23}}}\right]=\dfrac{2}{3}[u(x)]^{^{-\scriptsize\dfrac{1}{3}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(2x^3+x^2-6x)^{^{{\scriptsize\dfrac23}}} \\\\ &=\dfrac{d}{dx}\left[{u(x)}^{^{{\scriptsize\dfrac23}}}\right]&&\gray{\text{Let }u(x)=2x^3+x^2-6x} \\\\ &=\dfrac{2}{3}[u(x)]^{^{-\scriptsize\dfrac{1}{3}}}u'(x) \\\\ &=\dfrac{2}{3}[2x^3+x^2-6x]^{^{-\scriptsize\dfrac{1}{3}}}(6x^2+2x-6)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $g'( 2)$. $\begin{aligned} g'( 2)&=\dfrac{2}{3}\Bigl(2( 2)^3+( 2)^2-6( 2)\Bigr)^{^{-\scriptsize\dfrac{1}{3}}}\cdot\Bigl(6( 2)^2+2( 2)-6\Bigr) \\\\ &=\dfrac{2}{3}\cdot 8^{^{-\scriptsize\dfrac{1}{3}}}\cdot22 \\\\ &=\dfrac{2}{3}\cdot \dfrac{1}{2}\cdot 22 \\\\ &=\dfrac{22}{3} \end{aligned}$ In conclusion, $g'(2)=\dfrac{22}{3}$.